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CUET Mathematics 2025 PYQs15 May Shift 1. Free, no login required.

Calculus

past year questions, CUET Mathematics 2025 15 May Shift 1

Differential Equations

past year questions, Calculus, CUET Mathematics 2025 15 May Shift 1

Hard

core

For y≠0y \neq 0, the particular solution of the differential equation 2yex/ydx+(y−2xex/y)dy=02ye^{x/y}dx + (y - 2xe^{x/y})dy = 0 at the point (1, 1) is

Solution

✅ Correct Option: 1

We are given:

2yex/y dx+(y−2xex/y) dy=02ye^{x/y}\,dx + (y - 2xe^{x/y})\,dy = 0

Rearranging to express dxdy\frac{dx}{dy}:

dxdy=−(y−2xex/y)2yex/y\frac{dx}{dy} = -\frac{(y - 2xe^{x/y})}{2ye^{x/y}}

=2xex/y−y2yex/y= \frac{2xe^{x/y} - y}{2ye^{x/y}}

=xy−12ex/y= \frac{x}{y} - \frac{1}{2e^{x/y}}

The repeated appearance of x/yx/y suggests the substitution v=x/yv = x/y.

Let v=xyv = \dfrac{x}{y}, so x=vyx = vy.

dxdy=v+ydvdy\frac{dx}{dy} = v + y\frac{dv}{dy}

Substituting into the equation:

v+ydvdy=v−12evv + y\frac{dv}{dy} = v - \frac{1}{2e^{v}}

ydvdy=−12evy\frac{dv}{dy} = -\frac{1}{2e^{v}}

2ev dv=−dyy2e^{v}\,dv = -\frac{dy}{y}

∫2ev dv=−∫dyy\int 2e^{v}\,dv = -\int \frac{dy}{y}

2ev=−ln⁡∣y∣+C2e^{v} = -\ln|y| + C

Substituting back v=x/yv = x/y:

2ex/y+ln⁡∣y∣=C2e^{x/y} + \ln|y| = C

At x=1x = 1, y=1y = 1:

2e1/1+ln⁡∣1∣=C2e^{1/1} + \ln|1| = C

2e+0=C2e + 0 = C

C=2eC = 2e

2ex/y+ln⁡∣y∣=2e2e^{x/y} + \ln|y| = 2e

ln⁡∣y∣+2ex/y−2e=0\ln|y| + 2e^{x/y} - 2e = 0

ln⁡∣y∣+2(ex/y−e)=0\ln|y| + 2(e^{x/y} - e) = 0

Related questions:

2025: 14 May Shift 1

Consider the differential equation, xdydx=y(log⁡ey−log⁡ex+1)x\frac{dy}{dx} = y(\log_e y - \log_e x + 1), then which of the following are true?

(A) It is a linear differential equation

(B) It is a homogenous differential equation

(C) Its general solution is log⁡e(yx)=Cx\log_e\left(\frac{y}{x}\right) = Cx, where C is constant of integration

(D) Its general solution is log⁡e(xy)=Cy\log_e\left(\frac{x}{y}\right) = Cy, where C is constant of integration

(E) If y(1)=1y(1) = 1, then its particular solution is y=xy = x

Choose the correct answer from the options given below:

2026: 21 May Shift 1

General solution of the Differential equation dydx+ay=emx\frac{dy}{dx} + ay = e^{mx} is

2026: 18 May Shift 1

Match List-I with List-II

List-I Differential equationList-II Curve satisfying the differential equation
(A) d2ydx2+y=0\frac{d^2y}{dx^2}+y=0(I) x2+y2=Cx^2+y^2=C: Where C is an arbitrary constant
(B) dydx+y=x2+yx\frac{dy}{dx}+y=\frac{x^2+y}{x}(II) 3e−4y+4e3x=C3e^{-4y}+4e^{3x}=C: Where C is an arbitrary constant
(C) dydx+xy=0\frac{dy}{dx}+\frac{x}{y}=0(III) y=sin⁡x+cos⁡x+Cy=\sin x+\cos x+C: Where C is an arbitrary constant
(D) log⁡(dy/dx)=3x+4y\log(dy/dx)=3x+4y(IV) y=x+Cxe−xy=x+Cxe^{-x}: Where C is an arbitrary constant

Choose the correct answer from the options given below:

2026: 12 May Shift 2

Match List-I with List-II

List-I (Differential Equation)List-II (Order and degree)
(A) (d2ydx2)2+edydx=0\left(\dfrac{d^2y}{dx^2}\right)^2 + e^{\frac{dy}{dx}} = 0(I) order = 2, degree = 1
(B) xyd2ydx2+x(dydx)2−ydydx=0xy\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 - y\dfrac{dy}{dx} = 0(II) order = 2, degree = not defined
(C) (d3ydx3)2+log⁡e(dydx)=4x\left(\dfrac{d^3y}{dx^3}\right)^2 + \log_e\left(\dfrac{dy}{dx}\right) = 4x(III) order = 3, degree = 2
(D) ddx(d2ydx2)+ydydx=(1+dydx)1/2\dfrac{d}{dx}\left(\dfrac{d^2y}{dx^2}\right) + y\dfrac{dy}{dx} = \left(1+\dfrac{dy}{dx}\right)^{1/2}(IV) order = 3, degree = not defined

Choose the correct answer from the options given below:

2023: 30 May Shift 3

Solution of differential equation xdy−ydx=0x dy - y dx = 0 respresents :

2025: 3 June Shift 2

The general solution of the differential equation dydx=−4xy2\frac{dy}{dx} = -4xy^2 is given by

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